Express $z_1=3-3i$ in polar form. Express your answer in exact terms, using radians, where your angle is between $0$ and $2\pi$ radians, inclusive. $z_1=$
Solution: The Strategy A complex number in rectangular form, $z={a}+{b}i$, can be written in polar form as $z={r}[\cos{\theta}+i\sin{\theta}]$, where ${r}$ is the absolute value, or modulus, and ${\theta}$ is the angle, or argument. Therefore, ${r}$ and ${\theta}$ can be found using the following formulas: ${r}=\sqrt{{a}^2+{b}^2}$ $\tan{\theta}=\dfrac{{b}}{{a}}$ [How did we get these equations?] Similarly, a complex number in polar form, $z={r}[\cos{\theta}+i\sin{\theta}]$, can be written in rectangular form as $z={a}+{b}i$, using the following formulas: ${a}={r}\cos{\theta}$ ${b}={r}\sin{\theta}$ [How did we get these equations?] Finding $r$ For $z_1={3}{-3}i$ : ${a} = {3}$ ${b} = {-3}$ Therefore, we can find ${r}$ as follows. $\begin{aligned}{r}&=\sqrt{{a}^2+{b}^2} \\\\&=\sqrt{{3}^2+({-3})^2} \\\\&=\sqrt{9+9} \\\\&={\sqrt{18}} \\\\&={3\sqrt{2}}\end{aligned}$ Finding $\theta$ Using the formula, we have: $\begin{aligned}{\theta}&=\arctan\left(\dfrac{{b}}{{a}}\right) \\\\&=\arctan\left(\dfrac{{-{3}}}{{3}}\right) \\\\&={-\dfrac{\pi}{4}}\end{aligned}$ Since ${a}$ is positive and ${b}$ is negative, ${\theta}$ must lie in Quadrant $\text{IV}$. Therefore its angle must be between $\dfrac{3\pi}{2}$ and $2\pi$ radians. Using the identity $\tan(2\pi+\theta)=\tan(\theta)$, we know that the following is also a solution of the equation. $2\pi-\dfrac{\pi}{4}=\dfrac{7\pi}{4}$ So $\theta = {\dfrac{7\pi}{4}}$. Summary $z_1={3\sqrt{2}}\left[\cos{\dfrac{7\pi}{4}}+i\sin{\dfrac{7\pi}{4}}\right]$